What is the mass of lithium oxide formed by the interaction of 35 g of lithium with oxygen?

The oxidation reaction of lithium with oxygen is described by the following chemical reaction equation:

Li + ½ O2 = LiO;

When 1 mole of lithium reacts with 0.5 mole of oxygen, 1 mole of lithium oxide is formed.

Let’s find the amount of a substance contained in 35 grams of lithium.

To do this, we divide the mass of lithium by its molar mass.

M Li = 7 grams / mol;

N Li = 35/7 = 5 mol;

The same amount of lithium oxide substance will be obtained. Let’s find its mass by multiplying the amount of a substance by its molar mass.

M LiO = 7 + 16 = 23 grams / mol;

The mass of lithium will be:

m LiO = 5 x 23 = 115 grams;